1. Holt Algebra 2 5-3 Solving Quadratic Equations by Graphing and Factoring A zero (root) of a function is the x-intercept of the graph. Quadratic functions can have 0, 1, or 2 zeros. (In general, a function can have as many zeros as its highest exponent.) The zeros of a quadratic function are always symmetric about the axis of symmetry. Zeroes can be found by graphing or by factoring. No zeros:1 zero:2 zeros:

2. Holt Algebra 2 5-3 Solving Quadratic Equations by Graphing and Factoring Factoring by GCF The GCF (Greatest Common Factor) is the greatest number and/or variable that evenly divides into each term. Factor each expression by GCF: 1) 4xy 2 – 3x2) 10x 2 y 3 – 20xy 2 – 5xy 3) 3n 4 + 6m 2 n 3 – 12nm4) 5x 2 + 7 x(4y – 3x)5xy(2xy 2 – 4y – 1) 3n(n 3 + 2m 2 n 2 – 6m)Prime

3. Holt Algebra 2 5-3 Solving Quadratic Equations by Graphing and Factoring Determine the zeros of each function: 1)f(x) = 5x 2 + 10x 2)g(x) = ½x 2 – 2x 3)h(x) = 9x 2 + 3x 5x(x + 2) 5x = 0 and x + 2 = 0 x = 0 and x = -2 The zeros are x = 0 and x = -2 ½x(x – 4) ½x = 0 and x – 4 = 0 x = 0 and x = 4 The zeros are x = 0 and x = 4 3x(3x + 1) 3x = 0 and 3x + 1 = 0 x = 0 and 3x = -1 x = -1/3 The zeros are x = 0 and x = -1/3

4. Holt Algebra 2 5-3 Solving Quadratic Equations by Graphing and Factoring A binomial (quadratic expression with two terms) consisting of two perfect squares can be factored using a method called “difference of squares.” Difference of squares: a 2 – b 2 = (a + b)(a – b) Ex) Factor each expression: 1) x 2 – 9 2) 16x 2 – 49 (x + 3)(x – 3)(4x + 7)(4x – 7)

5. Holt Algebra 2 5-3 Solving Quadratic Equations by Graphing and Factoring Find the roots of the equation by factoring. Example 4A: Find Roots by Using Special Factors 12x 2 - 27 3(4x 2 – 9) 3(2x – 3)(2x + 3) = 0 2x – 3 = 0 2x + 3 = 0 x = 3/2 x = -3/2 The zeros are x = 3/2 and x = -3/2

6. Holt Algebra 2 5-3 Solving Quadratic Equations by Graphing and Factoring Factor by grouping when you have four terms with no common factor. Ex) Factor each expression: 1)xy – 5y – 2x + 10 2) x 2 + 4x – x – 4 y(x – 5) – 2(x – 5) (y – 2)(x – 5) x(x + 4) – 1(x + 4) (x – 1)(x + 4) x – 1 = 0 x + 4 = 0 x = 1 and x = -4 The zeros are x = 1 and x = -4